$h(t)=t^2+4t+3$ 1) What are the zeros of the function? Write the smaller $t$ first, and the larger $t$ second. $\text{smaller }t=$
To find the zeros of the function, we need to solve the equation $h(t)=0$. We can do that by factoring $h(t)$. $\begin{aligned} t^2+4t+3&=0 \\\\ (t+3)(t+1)&=0 \\\\ t+3=0&\text{ or }t+1=0 \\\\ t={-3}&\text{ or }t={-1} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $t$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }t\text{-coordinate}&=\dfrac{({-3})+({-1})}{2} \\\\ &={-2} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $h({-2})$ : $\begin{aligned} h({-2})&=({-2})^2+4({-2})+3 \\\\ &=4-8+3 \\\\ &=-1 \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }t&=-3 \\\\ \text{larger }t&=-1 \end{aligned}$ The vertex of the parabola is at $(-2,-1)$